chem 373- lecture 11: harmonic oscillator-i

8/3/2019 Chem 373- Lecture 11: Harmonic oscillator-I

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Lecture 11: Harmonic oscillator-I.

Vibrational motion

(This lecture introduces the classical harmonic oscillator

as an introduction to section 12.4 .

Lecture on-line

The Harmonic Oscillator-I (PDF)The Harmonic Oscillator-I (PowerPoint

Handout for this lecture (PDF)

Supporting material for classical harmonic oscillator (PDF)Supporting material for classical harmonic oscillator

(PDF handout format with 6 slides per page)

Supporting material for quantum mechanical harmonic oscillator (PDF)Supporting material for quantum mechanical harmonic oscillator (PDF

handout format with 6 slides per page)


2/19

Tutorials on-line

Basic concepts

Observables are Operators -

Postulates of Quantum Mechanics

Expectation Values -More Postulates

Forming Operators

Hermitian Operators

Dirac Notation

Use of Matricies

Basic math background

Differential EquationsOperator Algebra

Eigenvalue Equations

Extensive account of Operators


3/19

Audio-visuals on-line

Overview of the harmonic oscillator (PDF)(Good overview from the Wilson

group,****)

Overview of the harmonic oscillator (Powerpoint)

(Good overview from the Wilsongroup,****)

Vibrating molecule-I (Quick Time movie 1.4 MB)

(From the CD included in Atkins

,***)

Vibrating molecule-II (Quick Time movie 1.4 MB)

(From the CD included in Atkins ,***)

Slides from the text book (From the CD included in Atkins ,**)

The material in this lecture covers the following in Atkins.


4/19

Review : Classical harmonic oscillatorLet us consider a particle of mass m attached to a spring

Equilibrium

x=0,t=0

o

x

Stretchx=x

o

compressx=-xoxo

xo

At the beginning at t = o the particle is at equilibrium,that is no force is working at it , F = 0,


5/19

In general F = -k x . The force propotional to

displacement and pointing in opposite directiono

Equilibriumx=0,F=0

o

x

xo

xo

xo F=-kxo

xo

F= kxo

Review : Classical harmonic oscillator

k is the force constant

of the spring

V x k x( ) =1

2

2

F dVd x

dd x

k x k x= = = ( )22


6/19

We might consider as an other example two particles

attached to each side of a spring

re

A B

F= 0 Equilibrium

r = re

Case I: Equilibrium



7/19

re-x

A B

F= -k(-x) Equilibrium

r = re

Case III: Compress

x

A B

F= -kx Stretchr = re+x

Case II: Stretch

r = re+x

Again we have thatthe force F is proportional

to the displacementx and pointing in the

opposite directionF = - k x



8/19


x = A sin (k

mt )

V x k x( ) =1

2

2

FdV

d x

d

d x

kx k x= = = ( )2 2


9/19

V(x) = 1/2k2x2V

x-A2 A2k1 > k2

E

V(x) = 1/2k1x2

-A1 A1

k k1 2>

Appendix : Classical harmonic oscillator

The parabolic potential energy V =1/2

kx2 a harmonic oscillator, where x isthe displacement from equilibrium. Thenarrowness of the curve depends onthe force constant k: the larger the

value of k, the narrower the well.


10/19

Harmonic oscillator...Quantum mechanically

We shall now turn to a quantummechnical treatment of the one

dimensional harmonic oscillator

We have

H = E + Ekin pot

Hm

d

dxkx

E kxpot

= +

=

h2 2

22

2

2

1

2

1

2

V x kx( ) =

1

2

2Mass

Displacement

Force constant


11/19

The parabolic potential energy V =1/2

kx2 a harmonic oscillator, where x isthe displacement from equilibrium. Thenarrowness of the curve depends onthe force constant k: the larger the

value of k, the narrower the well.


V(x) = 1/2k2x2V

x-A2 A2k1 > k2

E

V(x) = 1/2k1x2

-A1 A1


12/19


=cycles per time unit

1

2

k

m

or

k m= 4 2 2 Thus

d

dx

d

dxmx

ddx

m x

d

dxx

m

H = -2m

kx = -2m

= -2m

(

= - 2m

where =

2 2 2

2

2

2

h h

h

h

h

h

2

2

2

22 2 2

2

2

2 2 22

2

2 2 2

1

22

4

2

+ +

)

( )


13/19


We must solve

H (x) =E (x)

or

-2m

(x) = E (x)

Thus

(x)(x) = - (x)

(x)(x)=0

2

2

2

( )

( )

h

h

h

d

dxx

d

dxx

mE

d

dx

mEx

2

22 2

2

22 2

2

22 2

2

2

+


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ddx

mE x

x f x

22

2 2

2

2

2

(x) (x) = 0

Let us look at a solution of

the form(x) = exp(-

2+ ( )

) ( )

h

let us further try to obtain

a power expansion of f(x)

of the form

f(x) = cnm=0

n= xn


15/19


The x f x

with f x xn

solution

=

(x) = exp(-

c

nm=0

n=2

2 ) ( )

( )

Must satisfy (- ) = ( ) = 0.

We

d

dx

mEx

want to solve :

2

22 22

(x)(x)=0

2+ ( )

h

It thi

E v

is shown in supplementarymaterial that s is only possible if

v = 1,2,3,4...= +h( )1

2 =k

m


16/19

E

1

2h

3

2h

5

2h

72h

x

9

2h

112

h

The energylevels of a

harmonic

oscillator

are evenlyspaced with

separation

, with =

(k/m)1/2.

Even in its

lowest state,

an oscillatorhas an

energy

greater

than zero.

Harmonic oscillator...Quantum mechanically .Energy levels

v = 0

v = 1

v = 2

v = 3

v = 4

v = 5

v = 6E v= +h( )1

2


17/19

We obtain as the solutions to (x)

Gaussian function)

( ) ( ) (

( ) exp ( )

x N polynomial in x bell shaped

x Ny

H yv v v

=

=

2

2

Harmonic oscillator...Quantum mechanically.... Wavefunction

With

mk =h

2 1 4

/y x= /

Where

Nv = 1

2

1

2 v v!


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_________________________

v H

1

1 2y

2 4y - 23 8y - 12y

4 16y - 48y +12

5 32y -160y +120y6 64y - 48y + 72y - 120

_____________________________

v

2

3

4 2

5 3

6 4 2

0

Hermit polynominals



19/19

What you should learn from this lecture

1

12

12

2

.

) ) .

)

The potential energy in a harmonic oscillator is given

V( x) = V(x x k(x

k is the force constant and x is the position where

the force F = -dV

dx

(x is zero

e 2 e

e

e

= =

=

x k x

Here

k x

2

12

.

( )

You are not required to solve the Schrdinger eq.For the quantum mechanical harmonicoscillator. However you should recall that the energy is given by :

v = 1,2,3,4...E v= +h = km

3. Also recall for the solution (x)

=

v v vx N

yH y y x

mk( ) exp ( ); / ;

/

=

=

2 2 1 4

2

hWhere

Nv =1

2

1

2 vv!

chem 373- lecture 11: harmonic oscillator-i

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